If the gas begins at the conditions described in Trial 1, determine the volume of the gas at STP.

A student measures the absorbance of a sample of red dye #3 using a spectrophotometer. If the absorbance is measured as 0.468, w

Gre4nikov [31]

Answer:

5.88 μM is the concentration of red dye in the sample.

Explanation:

Absorbance is defined s capacity of a substance to absorb a light of a specified wavelength.It is calculated by using Beer-Lambert Law:

$A=\epsilon lc$

A = absorbance

$\epsilon$ = Molar absorptivity

c = concentration

l = Length of the solution through which light passes

Given, A = 0.468, l = d = 1.00 cm,c = ?

$\epsilon =7.96\times 10^4 M^{-1} cm^{-1}$

$A=\epsilon lc$

$0.468=7.96\times 10^4 M^{-1} cm^{-1}\times 1.00 cm\times c$

$c=5.8793\times 10^{-6} mol/L=5.8793 mol/(\mu L)\approx 5.88mol/(\mu L)$

( $1 L= 1,000,000 \mu L$ )

5.88 μM is the concentration of red dye in the sample.

5 0

3 years ago

Which of the following factors would negatively impair the water quality in a watershed? I. irresponsible use of industrial chem

Deffense [45]

Answer:

1 and 2 only

Explanation: These will make a bad impact on water.

3 0

4 years ago

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Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

3 years ago

A student took notes in class. - Uses high frequency sound waves - Creates images from echoes - Has many applications - Medical

SpyIntel [72]

The correct option should be ultrasound technology (option B) because it is related to sonographers or ultrasound technicians. they are most likely with while pregnancy but they have plenty of uses, such as evaluating and diagnosis, many medical treatment for elderly patients.

5 0

3 years ago

Read 2 more answers

True or false carbon-15 and nitrogen-15 are isotopes of each other

Alla [95]

Answer:

I would say false

Explanation:

8 0

3 years ago